Bear H.S. Understanding Calculus

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Chapter 23 • Series

139

EXAMPLE

23.8

lOOn

L....J

+ I .

Here the dominant term in the numerator is n

and the dominant term in the denominator is 3n

•

Therefore, for large n, the fraction will behave like

-i;;,

so we make a limit comparison with L

n2+

lOOn

3n . 3n

+ 100n

. - = lim = I.

Il~OO

+ 1 1

n~oo

+ 1

The limit is nonzero, so both series behave in the same way. L

diverges so that the given series

diverges.

PROBLEMS

Find the following finite sums. Brute force with a calculator will work, but a little thought will

save time and be more interesting.

23.1 L 100

n=1

100

23.2

Ln.

Hint: If S =

L:'~1

then also S =

L~~I(101-n).

Hence

S+S

L,I'~1101.

n=1

23.3

11=1

23.4

L(2n

- 1) = 1 + 3 + 5 +

...

+ 19

11=1

= (1 +

19)+(3+17)+(5+

15)+(7+

13)+(9+

11)

23.5

L(2n

- 1) = 1 + 3 + 5 +

...

+ t2N - 1)

11=1

23.6

,,~

,,=1

23.7

tGr

n=2

Show that the series converges, or show that it diverges.

23.8

L(_I)n_n_.

n+l

23.9

L(-lt~

23.10

L(-l)n

logn

n+l

23.11

"<_I)n_

140

Understanding Calculus

23.12 L:-

Iogn

23.13

logn

23.14

L:e-

23.15

".!!-

L...J

23.16

23.17

(n + 1)Iogn

23.18

"-2

L...J

1 1

23.19 " - sin

Hint: Comparewith

L...J

2320

· .

I· 4"

•

--.

Int.

Imn~oo

n3"

n·3

23.21

L:e~

23.22

3" + 4"

23.23 Showthat if

2:0 for all

nand

converges, then L a; converges.

23.24 Showthat if

2: 0 for all

nand

converges, then L sin

converges.

Power

Series

Series of the form

Lan(x - XO)'I

n=O

(24.1)

are called

power series. The sum of a power series is a function of x, and all our elementary

functions (log

x, e", sin

JX,

cos-

x, rational functions, etc.) can be expressed locally

as power series. The two types

series in (24.1), one in powers of x, and one in powers

(x - xo), are obviously similar, and any statement about one type of series implies a

corresponding statement about the other kind. For example, if

L a

nx"

converges for all x

such that

I < 2, then L an(x - 3)n converges for all x such that

< 2, and vice versa.

We will therefore make our calculations with series in powers of

x for simplicity; we realize

that all our results will apply with appropriate changes to series in powers of

(x - xo).

The power series L a

nx"

converges or diverges depending on the value of x. Since the

terms

lanx

Iget larger or smaller as [xIgets larger or smaller, we expect that the closeness

x to the origin is the key to convergence. It is, and here is the precise statement:

Convergence Condition: If

limll~oo

anx

= 0 then L

a.x"

converges absolutely if

[x] <

Ixol·

Ifanx

-Pf

0, then

Lanx"

diverges iflxl 2:

Ixol.

To verify the condition. first notice that if anx

-Pf

0, then anx

-Pf

0 for all [xI 2: Ixo

so the series surely diverges at

and all [xI 2:

Ixo

Now suppose allx

This does not

imply that

L anx

converges, but it does imply that lanx

I < 1 for all large n. If [x] <

Ixol,

so that

1..£

I < 1, then for large n,

lanxnl

lanx;;II~ln

I~I"

Since L

1*1

is a convergent geometric series, L lanx

I converges if [xI <

Ixo

I; that is,

L anx

converges absolutely if

I <

Ixo

I·

We saw in Chapter 21 that Inlx" I

for all x

Hence, L n!x" converges only

x =

Such a series is of no use, since we want to use series to represent functions, and we

will therefore now consider only series that converge for some nonzero values of

141

142

Understanding Calculus

If a power series converges for some values of x

:j::.

0 and diverges for some x, then it

follows from the ConvergenceCondition that there will either be a largest number r such that

a.r"

0, or a snlallest number r such that

a.r"

-A

In either case, the series will converge

absolutelyif

I < r and divergeif

I > r. The numberr is called the radius of convergence,

and the interval

(-r,

is the interval of convergence. The series may convergeat either or

neither or both of the numbers

rand

-r.

The convergence or divergence at these points is

generally of no importance.

asx"

0 for all x, then the series converges absolutely for all x, and we say the

radius of convergenceis

and the interval of convergence is the whole line

(-00,

(0).

For

example, we saw in Chapter

21 that

0 for all x, so L

converges absolutely for all

x. We will show in Chapter 26that

x" x2 x3 x

= L - = 1

+ - + - + - +

....

n=O n! 2! 3! 4!

Recallthat the sequences{(logn)k), {n

}

with p > 0, {an}witha > 1,and

{n!}

represent

different orders of growth in the sense that

(log

n)k 0 n" an

, -

0 -

(24.2)

an'

Since

0 for all a > 1, it follows, writing x = 1, that n''x"

0 for all

]r]

< 1.

a a

EXAMPLE

24.1

Find the radius of

convergence

of L

= L n

)n.

From (24.2) we know that n

2x"

0 if [r] < 1, and of course In

if [x]

1. Hence,

(1)"

0 if andonly if

111

< 1. The radiusof convergence is 3, and the series

converges

absolutely

(-3,3).

Coefficientsthat are powers of n, like n

in Example 24.1, have no effect on the radius

of convergence. All three of the following series have the same radius, 3:

This follows from the convergencecondition because both

(t)nIand I

)Iapproach 0 if

ItI< 1, and both tend to

if IJI > 1. A similar statementholds for logn or powers of log n.

For example, both I(logn)kXU I and I

(10;

n )k x" I tend to zero if IxI < 1, and both tend to

I > 1. Therefore, all three of the following series have radius 1 for any value of k:

" (logn)kx", " x"

1 x"

LJ LJ

(log n)!

EXAMPLE

24.2

Find the radiusof L 2"(logn)3X" and L

(l0~/~)3

x",

Wewritethe terms in the following form: (logn)3(2x)n and (10;11)3 (2x)n. Both terms approachzero for

12x

I < 1,and both tend to

12x

I > 1. Therefore, the radius for both series is

The following test, the ratio test, is a popular technique because it requires very little

thought. The ratio test applies to any series, so we state the test for a series of constants,

L an,

and then show how it applies to power series.

Ratio Test:

The senes S"

converges absolutely

iflimll~oo

an+1

I < 1 and diverges

L..J an

lim

--.

a~:1

I > 1.

Chapter 24 • Power Series 143

To verify the ratio test, first notice that if lim

H x

)

Ian+1 I > 1, then la

Ian

Ifor all

large n, so

-r;

0 and the series surely diverges. Now assume

limn~oo

+1 I = r < 1, and

let s be a number between

rand

1. Then Ian+1 I < s for all sufficiently large n - say for all

N. Hence,

a;;

1 1 <

1 < s,

aN+l

and in general

laN+k

I < IaNIs

Since L sk and L IaNIs

are convergent geometric series,

Ian

Iconverges.

EXAMPLE

24.3

Use the ratio test to findthe radius of convergence of L

n~~~~2

x".

log(n + I) n

1 I

lim x

•

---

n-+-oo

(n + 1)25

logn x

= [xl lim

_n_

2_log(n

+ I)

n-+-oo

(n + 1)2 logn 5 5

The series convergesif I

I < 1; the radius is 5.

Example 24.3 could also be done simply by rewriting the terms:

log

n n log n

(X)

n25n+2

x = 25n

The factor

does not change the radius from that of L (

j". The ratio test provides a

mechanical way of showing the insignificance of

relative to

(~)n.

The next example shows a more realistic use of the ratio test.

EXAMPLE

24.4

Find the radius of convergence of L

~xn.

Wecalculate the limitingratio:

. I (n + I)! n+l nn 1 I

lim

x·--

n-+-oo

(n + I)n+l n! x

. (n + I)! n"

Ixl lim

---

n-+-oo

n! (n + l)n+l

(n+l).n!

[r]

hm .

------

n-+-oo

n! (n + 1) . (n + l )"

· (

n ) n

1·

1 IxI

x 1m

= [x]

Hfl

n-+-oo

n + 1

n-+-oo

(I + *)n e

The radius is

From the preceding examples we see that for power series L a.x", the ratio test always

looks like this:

lim

an+1xn+11

= [x] lim Ian+l I·

n~oo

anx

n~oo

The radius is the reciprocal of

limn~oo

Ian+1

144

Understanding Calculus

A power series represents a function on its interval of convergence, and much of the

importance

power series derives from the fact that they can be differentiated and integrated

term-by-term. That is, if

I(x)

Lan

=ao +

alx

a2x2

...

n=O

and the series converges on

(-r,

r), then I'(X) and f;

I(t)

dt exist on

(-r,

r), and

(x) = L nanx

= al + 2a2x +

3a3x2

...

n=1

(24.3)

(24.4)

(24.5)

n+1 al 2 a: 3

f(t)dt

= L

--x

aOx

-x

....

n=O

n + 1 2 3

The

radius does not change when you differentiate or integrate a series term-by-term, and the

resulting series represent the right functions. If

I (x) has a power series representation, then

I (x) has derivatives

all orders, and all these derivatives have power series, and all the series

have the same radius

convergence.

Now consider some

the functions we can represent with series, starting with the

geometric series. If

I < 1, then

'""

n 2 3

--=L...JX

=1+x+x

+ ....

I-x

n=O

(24.6)

Since

I - x I < 1 and I ± x

1 < 1 if and only if [x I < 1, we can substitute in (24.6) to get

~()n

1 2 3 4

L...J

-x

= - x + x - x + x -

...

1 + x

n=O

= '"'"'(x

2)n

= 1

+ ... ,

1 - x

L..J

n=O

(24.7)

(24.8)

_1_

= '"'"'(_x2)n = 1 _ x

+ x

+ ... . (24.9)

+ x

L..J

n=O

Now the series above can be differentiated or integrated, and for x in

(-1,

1) we get:

dx 1 2 1 3 1 4

10g(1 + x) = 1 + x = x - 2

+ 3

+ , (24.10)

_1_

= 1 + 2x + 3x

+ 4x

+ (24.11)

(l-x)2

I-x

-I

f dx 1 3 1 5 1 7

tan x = 1 +x

= X -

3"x

+ s-X

-"7

+....

(24.12)

The

series for 10g(1 + x) and

tan-

x are proper alternating series when x = 1, and it can be

shown that they converge to the right thing; that is,

1 1 1

log2=1-

+ ··· , (24.13)

1 n 1 1 1

tan-

1 = - = 1 - - + - - -

+...

. (24.14)

357

Chapter24 • PowerSeries

145

If we substitute 3x for x in the series for 1

~x'

we get

_1_

= 1 + (3x) + (3x)2 + (3x)3 +

...

, (24.15)

and this series converges provided

13x

I < 1; that is, provided [xI <

Similarly, if we replace

x by

in (24.11), we get

1 4

2x 3x

=1+-

+-4

+-8

...

(1-

~)2

-x)2

and this series converges if I

I < 1, or

I < 2.

EXAMPLE

24.5

Find a powerseriesfor

10g(1

+3x) and findits radiusof

convergence.

From (24.10)we have,for IxI < 1,

2 1 3 1 4

10g(1

+ x) = x -

-x

+ - x -

-x

....

2 3 4

Wecan substitute 3x for x as long as

13x

I < 1,or IxI <

Therefore,

10g(1

+ 3x) = 3x -

-(3x)

...

234

and the radiusof convergence is

EXAMPLE

24.6

Wewill showin Chapter26 that

. 1

Slnx

=x -

-x

+....

3! 5! 7!

Use this to finda seriesfor cosx. What is the intervalof convergence for both series?

Since

sinx = cosx, we have

d ( x

)

cosx = dx x - 3! + 5! - 7! +

...

=1--+---+···.

2! 4! 6!

'"'"

n X

LJ(-I)

n=O (2n)!

(O!

is definedto be 1.) Tofindthe intervalof convergence, we use the ratio test:

. I X

(2n)! I

2·

lim .

I lim =

n--+oo (2n + 2)! x

n--+oo (2n + 2)(2n + 1)

The limitingratio is less than 1 for all x, so the seriesconverges for all x.

EXAMPLE

24.7

Find a powerseriesfor cos

Sincethe seriesfor cosx converges for all x, we can substitute

for x:

cos

= 1 _

(~)2

(~)4

(~)6

+ ....

2 2! 2 4! 2 6! 2

(24.16)

146

Understanding Calculus

PROBLEMS

Find the radiusof

convergence.

24.1

24.2

(2x )n

24.3

(~r

.n 4

24.4 L 2

(logn)x

24.5

"~xn

L..J

24.6 L n +

.;n

x·

logn

24.7

"~Xn

L..J

24.8

,,_n

L..J

24.9 L n!x·

(2n)!

24.10

Lenx

24.11

,,~(~)n

L..J

+ I 2

'+ 2

24.12 L n. n x

n! +logn

.24.13

"n!

+ IOnx

L..J

n!.

IOn

24.14

(logn)k

x·

(k and p are arbitrary numbers.)

L...J

Find the intervalof

convergence.

24.15 L n

(x - 3)n

24.16 L

-(x

- l)n

24.17 "

-(x

5)n

L..J

24.18 "

-(x

+ I)n

L...J

Use the seriesgivenin the chapterfor e", sinx, cosx, and the series derivedfrom the seriesfor

I~x'

to findthe series for the givenfunction by substitution, differentiation, or integration. Give

the intervalof convergence.

24.19 e

24.20

24.21

10g(1

24.22

10g(1

2x)

Chapter24 • PowerSeries

24.23

(1 +

X)2

24.24

I +x

24.25 sin2x

24.26 tan"

24.27 cos

24.28 (1 +

X)2

147

24.29

sinx

24.30 x

1+x

24.31

log(1 +x

)

24.32 Showthatif

limn~oo

+ \ I =

so r is theradiusof convergence of

a.x", thenthe ratio

an r

L-

test givesthe same radius for L

na.x'""

and L

n~l

•

Hint: For x

0, L na.x":"

converges

if and only if L na.x"

converges,

and L

n~1

converges

if and only if

..E!L

converges.

x+l

(25.1)

Taylor

Polynomials

In this chapter, we show how functions can usefully be approximated by polynomials. For

example, there is no elementary antiderivative for

«', so Jo

x 2

dx can only be evaluated by

some approximation method. If we can find (and we can) a polynomial

P(x)

that is close to

x 2

on [0, I], then Jo

P(x)

approximatesJo

x 2

dx.

In the next chapter, we use these ideas

to show how functions can be represented by power series.

Suppose we want to find an nth degree polynomial that approximates a function

f(x)

as well as possible near zero. The reasonable thing to do is to find the polynomialthat agrees

with

f (x) at 0 and has the same derivatives as f (x) at

All the derivatives of an nth degree

polynomial are zero beyond the nth derivative, so we can only match the first

n derivatives of

f(x)

at 0 with those of an nth degree polynomial. If

Pn(x) =ao +

alx

+ a2x2 +

...

+ a.x",

then P;(x) has the following derivatives:

(x)

+ 2a2x + 3a3x2 + ... + na.x":",

P~'

(x)

= 2a2 + 3 . 2a3x + 4 . 3a4x2 + ... +

n(n

l)a

P~"(x)

3·

2a3 +

4·3·

2a4x +

5·4·

3a5x2 + ... + n(n -

I)(n

- 2)a

p~n)(x)

=n!a

Evaluating the derivativesat x =0, we get

Pn(O)

= ao,

P~(O)

aI,

P~'(O)

= 2a2,

P~"(O)

3·

2a3,

and in general

p~k)(O)

= kuu; 0

To match up the derivativesof Pn(x) with those of

f(x)

at 0, we must have for each k

p~k)(O)

= kia; = f(k)(O),

j(k)(O).

149